Buds Rehanging solution codeforces

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A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. The parent of a vertex vv (different from root) is the previous to vv vertex on the shortest path from the root to the vertex vv. Children of the vertex vv are all vertices for which vv is the parent.
A vertex is a leaf if it has no children. We call a vertex a bud, if the following three conditions are satisfied:
 it is not a root,
 it has at least one child, and
 all its children are leaves.
You are given a rooted tree with nn vertices. The vertex 11 is the root. In one operation you can choose any bud with all its children (they are leaves) and rehang them to any other vertex of the tree. By doing that you delete the edge connecting the bud and its parent and add an edge between the bud and the chosen vertex of the tree. The chosen vertex cannot be the bud itself or any of its children. All children of the bud stay connected to the bud.
What is the minimum number of leaves it is possible to get if you can make any number of the abovementioned operations (possibly zero)?
The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. Description of the test cases follows.
The first line of each test case contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of the vertices in the given tree.
Each of the next n−1n−1 lines contains two integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) meaning that there is an edge between vertices uu and vv in the tree.
It is guaranteed that the given graph is a tree.
It is guaranteed that the sum of nn over all test cases doesn’t exceed 2⋅1052⋅105.
For each test case print a single integer — the minimal number of leaves that is possible to get after some operations.
input
5 7 1 2 1 3 1 4 2 5 2 6 4 7 6 1 2 1 3 2 4 2 5 3 6 2 1 2 7 7 3 1 5 1 3 4 6 4 7 2 1 6 2 1 2 3 4 5 3 4 3 6
output
2 2 1 2 1
In the first test case the tree looks as follows:
In the second test case the tree looks as follows:

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